# Thread: Function of a Function Quotient Integration

1. ## Function of a Function Quotient Integration

Hi All please can i have some help with this!

$

\int {\frac{{e^{\tan ^{ - 1} 3x} }}
{{1 - 9x^2 }}}

$

2. Originally Posted by MexicanGringo
Hi All please can i have some help with this!

.......arctan3x
y = e..................
..........1 + 9x(squared)

Apologies for poor formula havn't figured how to do it properly yet
note that the derivative of arctan 3x is 1/(1 + 9x^2). thus, a substitution of u = arctan 3x will yield int e^u du. i trust you can take it from here?

EDIT: did i miss something? is LaTeX not working?

3. You must have been gone for awhile JH. No, Latex is not working and hasn't been for sometime now.

4. Thanks for the help JHevon, OK not sure if this is correct as the type of problem is the inverse of the usuall f'(x) / f(x)

y = e^arctan3x
......1+9x^2

u = arctan3x

du/3 = dx/1+9x^2

= e^arctan3x/3 + C

Is this correct?

5. Hello, MexicanGringo!

$\int \frac{e^{\arctan3x}}{1+9x^2}\,dx$

We have: . $\int e^{\arctan3x}\cdot\frac{dx}{1+9x^2}$

Let: $u \:=\:\arctan3x \quad\Rightarrow\quad du \:=\: \frac{3\,dx}{1 + 9x^2} \quad\Rightarrow\quad \frac{dx}{1+9x^2} \:=\:\frac{1}{3}\,du$

Substitute: . $\int e^u \cdot \frac{1}{3}\,du \;=\;\frac{1}{3}\int e^u\,du \;=\;\frac{1}{3}\,e^u + C$

Back-substitute: . $\frac{1}{3}\,e^{\arctan3x} + C$

6. Excellent!! Thank you all for your help!!